Title Med-RM_Phy_SP-3_Ch-17_Electrostatic Potential and Capacitance.pdf ✅

Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph. 011-47623456 Chapter Contents Introduction Electric energy can be stored in a common device called a capacitor, which is found in nearly all electronic circuits. A capacitor is used as a storehouse for energy. Capacitors store the energy in common photoflash units. The rapid release of this energy is evident in the short duration of the flash of camera used to take photographs. Introduction Electrostatic Potential Energy Electrostatic Potential Potential due to a Point Charge Potential due to System of Charges Potential due to an Electric Dipole Equipotential Surfaces Potential Energy in an External Field Electrostatics of Conductors Capacitors and Capacitance Combination of Capacitors Van de Graaff Generator ELECTROSTATIC POTENTIAL ENERGY Potential energy of a configuration of charges is the work done by an external force to bring the charges from infinity to their respective position without acceleration. (1) 1 2 0 1 4 q q U r   1 2 1 2 if 0, 0 if 0, 0 qq U qq U           r q1 q2 1 2 2 0 1 4 dU q q F dr r    [Coulomb’s law] (2) Potential energy for a system of more than two charges For a system of more than two charges, the total potential energy can be calculated by making all possible combination of two charges at a time, and computing their interaction potential energy. For example, for a system of three charges, the potential energy is U = U12 + U13 + U23 Chapter 17 Electrostatic Potential and Capacitance
128 Electrostatic Potential and Capacitance NEET Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph. 011-47623456 Q Q Q 2r 2r 2r Q Q r Q r r (changed to) 2 1 0 3 4 Q U r    2 2 0 3 4 (2 ) Q U r    Wext = U2 – U1 = 2 2 0 0 3 11 3 4 2 4 (2 ) Q Q rr r            Welectric field = 2 1 2 ext 0 3 4 (2 ) Q UU W r     (3) Electrostatic potential energy of system of four charges as shown in figure. Q Q Q a a Q a 2 U = U12 + U13 + U23 + U14 + U24 + U34 2 2 0 0 4 2 4 4 ( 2) Q Q U a a     2 0 (4 2) 4 Q a    (4) Energy required to move Q to infinity is 0 0 1 1 4 4 Qq Qq r r    r r q(fixed) Q q(fixed) 2r Þ  2 0 0 12 4 24 4 Qq Qq mv v r rm       . This is escape velocity. Example 1 : A charge is moved in an electric field of a fixed charge distribution from point A to another point B slowly. The work done by external agent in doing so is 100 J. What is the change in potential energy of the charge as it moves from A to B? What is the work done by the electric field of the charge distribution as the charge moves from A to B? Solution : Wext = U = UB – UA = 100 J As, Fext = – FE for the charge to move slowly, so WE = – Wext = – 100 J.
NEET Electrostatic Potential and Capacitance 129 Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph. 011-47623456 ELECTROSTATIC POTENTIAL Work done by an external force in bringing a unit positive charge from infinity to a point in electrostatic field of some another charge without acceleration is equal to electrostatic potential (V) at that point. Electrostatic potential energy Electrostatic potential = Charge It is a scalar quantity. Unit for Electrostatic Potential The unit of measurement for electrostatic potential is the volt, so electrostatic potential is often called voltage. A potential of 1 volt equals 1 joule of energy per 1 coulomb of charge. joule 1 volt = 1 coulomb Electrostatic Potential Difference Similar to electrostatic potential, the electrostatic potential difference is the work done by external force in bringing a unit positive charge from one point R to other point P without any acceleration. VP – VR = U U P R – q Note : As before, that it is not the actual value of potential but the potential difference that is physically significant. If, as before, we choose the potential to be zero at infinity, the above equation implies. Example 2 : Find the change in electric potential energy, U as a charge of (a) 2.20 × 10–6 C. (b) –1.10 × 10–6 C moves from a point A to a point B, given that the change in electric potential between these points is V = VB – VA = 24.0 V. Solution : (a) Solving V = U / q0 for U, we find U = q0V = (2.20 × 10–6 C) (24.0 V) = 5.28 × 10–5 J (b) Similarly, using q0 = –1.10 × 10–6 C, we obtain U = q0V = (–1.10 × 10–6 C) (24.0 V) = –2.64 × 10–5 J Example 3 : A particle of mass m and positive charge q is released from point A. Its speed is found to be v when it passes through a point B. Which of the two points is at higher potential? What is the potential difference between the points? Solution : The point A is at higher potential than B as the particle has gained kinetic energy while moving from A to B. The kinetic energy acquired by the particle is equal to loss in potential energy. i.e., K = q(VA – VB) 1 2 () 2  mv q V V   A B 2 1 – 2 A B mv V V q   Example 4 : A charge of 10 C is moved in an electric field of a fixed charge distribution from point A to another point B slowly. The work done by external agent in doing so is 100 J. What is the potential difference VA – VB ? Solution : VA – VB = U U A B – q = Wext q . Here, Wext is work done in moving the charge from B to A i.e., – 100 J  VA – VB = –10 volts.
130 Electrostatic Potential and Capacitance NEET Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph. 011-47623456 Example 5 : Suppose an electron in the picture tube of a television set is accelerated from rest through a potential difference Vb – Va = Vba = + 5000 V. (a) What is the change in electric potential energy of the electron? (b) What is the speed of the electron (m = 9.1 × 10–31 kg) as a result of this acceleration? Solution : The electron, accelerated towards the positive plate, will change in potential energy by an amount PE = – qVba . The loss in potential energy will equal its gain in kinetic energy (energy conservation). (a) The charge on an electron is q = – e = – 1.6 × 10–19 C. Therefore its change in potential energy is PE = – qVba = (– 1.6 × 10–19 C)(+5000 V) = – 8.0 × 10–16 J. The minus sign indicates that the potential energy decreases. The potential difference, Vba, has a positive sign since the final potential Vb is higher than the initial potential Va. Negative electrons are attracted towards a positive electrode and repelled away from a negative electrode. (b) The potential energy lost by the electrons becomes kinetic energy KE. From conservation of energy, KE + PE = 0, so  KE = – PE    1 2 –0 – – – 2 mv q V V qV   b a ba . where the initial energy is zero since we are given that the electron started from rest. We solve for V.     –19 –31 2 2 –1.6 10 C 5000 V – – 9.1 10 kg qVba V m     = 4.2 × 107 m/s. POTENTIAL DUE TO A POINT CHARGE Consider a point charge q placed at point O. Consider any point P in the field of the above charge. q r P E O The potential at point P due to charge q   q V r 4 0 or  Kq V r , where  0 1 4 K Use the value of charge with sign to calculate potential.